Signed area versus geometric area | Year 12

01 / LEARN

Understand the idea.

The integral keeps the sign

A definite integral records signed area: values above the axis contribute positively and values below contribute negatively. Contributions may cancel even when the graph encloses substantial regions. A zero integral therefore does not mean zero geometric area.

Geometric area counts each piece positively. Find the zeros within the interval and determine the sign between them. Reverse the sign of any negative integral before adding the pieces; use square units when both coordinates measure lengths.

The line y=x−1 crosses the x-axis at x=1; shown from x=0 to x=3.

Use a sign chart before integrating

Build a sign chart from a factored expression, a test point in each interval, or the supplied graph. Splitting at every zero is safe, although a zero where the sign does not change does not require a reversal.

The notation $\int_{a}^{b} ∣ f (x) ∣ d x$ describes geometric area, but hand calculation still needs the sign analysis. Taking the absolute value after adding all signed pieces loses regions that have already cancelled.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

Find both $\int_{0}^{3} (x - 1) d x$ and the area between the graph and axis.

Worked solution

1
Locate the sign change.
$x - 1 = 0 \Rightarrow x = 1$
The line is negative from 0 to 1 and positive from 1 to 3.
2
Find the signed integral over the whole interval.
$\begin{aligned} \int_{0}^{3} (x - 1) d x & = {[\frac{x^{2}}{2} - x]}_{0}^{3} \\ = \frac{9}{2} - 3 = \frac{3}{2} \end{aligned}$
Upper minus lower keeps the cancellation between negative and positive contributions.
3
Calculate the two signed pieces.
$\begin{aligned} I_{1} & = {[\frac{x^{2}}{2} - x]}_{0}^{1} = - \frac{1}{2} \\ I_{2} & = {[\frac{x^{2}}{2} - x]}_{1}^{3} = 2 \end{aligned}$
Splitting at the zero separates the below-axis region from the above-axis region.
4
Add magnitudes for geometric area.
$A = | - \frac{1}{2} | + ∣ 2 ∣ = \frac{5}{2}$
Geometric area counts both regions positively rather than cancelling them.

Answer

Signed integral $3 / 2$ ; area $5 / 2$ .

EXAMPLE 02

Worked example

Find the area between $y = x^{2} - 4$ and the axis for $- 2 \leq x \leq 2$ .

Worked solution

1
Identify the sign on the interval.
$x^{2} \leq 4 \Rightarrow x^{2} - 4 \leq 0 (- 2 \leq x \leq 2)$
The whole region lies on or below the axis, with zeros at the endpoints.
2
Reverse the integrand for area.
$A = \int_{- 2}^{2} (4 - x^{2}) d x$
Axis height minus curve height is nonnegative throughout the interval.
3
Evaluate both endpoint values.
$\begin{aligned} A & = {[4 x - \frac{x^{3}}{3}]}_{- 2}^{2} \\ = (8 - \frac{8}{3}) - (- 8 + \frac{8}{3}) \\ = \frac{32}{3} \end{aligned}$
Subtracting the negative lower endpoint value adds the equal left-hand region.

Answer

$32 / 3$

03 / YOUR TURN

Try it yourself.

Check your understanding

For $y = x$ on $[- 1, 1]$ , what are the signed integral and geometric area?

Show answer & reasoning

$0$ and $1$

The triangles on $[- 1, 0]$ and $[0, 1]$ each have area $\frac{1}{2}$ . Their signed contributions are opposite, giving integral 0, while their geometric areas add to 1.

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6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Find the area between $y = x - 2$ and the axis on $[0, 4]$ .
3 marks
Find $\int_{- 2}^{2} x d x$ and the corresponding geometric area.
3 marks
Find the area under $y = 4 - x^{2}$ on $[- 2, 2]$ .
3 marks
Find the area between $y = x (x - 2)$ and the axis on $[0, 3]$ .
3 marks
Find the area between $y = sin x$ and the axis on $[0, 2 π]$ .
3 marks
A student says an integral of 0 proves that an area is 0. Give a counterexample.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the area between $y = x - 2$ and the axis on $[0, 4]$ .
3 marks
Split at 2.
Two triangles each have area 2.
$4$
Find $\int_{- 2}^{2} x d x$ and the corresponding geometric area.
3 marks
The signed contributions are $- 2$ and 2.
Their absolute values sum to 4.
$0$ and $4$
Find the area under $y = 4 - x^{2}$ on $[- 2, 2]$ .
3 marks
The function is nonnegative, so integrate it directly.
$\frac{32}{3}$
Find the area between $y = x (x - 2)$ and the axis on $[0, 3]$ .
3 marks
An antiderivative is $\frac{x^{3}}{3} - x^{2}$ .
Contributions on $[0, 2]$ and $[2, 3]$ are $- \frac{4}{3}$ and $\frac{4}{3}$ .
Add their magnitudes.
$\frac{8}{3}$
Find the area between $y = sin x$ and the axis on $[0, 2 π]$ .
3 marks
On $[0, π]$ the area is 2; on $[π, 2 π]$ the signed integral is $- 2$ .
Add 2 and 2.
$4$
A student says an integral of 0 proves that an area is 0. Give a counterexample.
3 marks
The integral is 0 but the two triangular regions have total area 1.
$f (x) = x$ on $[- 1, 1]$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods integration and logarithmic antiderivatives; original learning and assessment material, with by-hand methods and explicitly described numerical models.

Official curriculum source ↗

In this lesson

Understand the idea.

The integral keeps the sign

Use a sign chart before integrating

See the method in action.

Worked example

Locate the sign change.

Find the signed integral over the whole interval.

Calculate the two signed pieces.

Add magnitudes for geometric area.

Worked example

Identify the sign on the interval.

Reverse the integrand for area.

Evaluate both endpoint values.

Try it yourself.

Check your understanding

Keep practising